Kepler’s Third Law
As we have derived the equations of Orbital velocity of satellite in our last post, this is the right time to discuss on an important Law of Physics, which talks on the period of its revolution. This is known as Kepler’s Third Law or 3rd Law of Kepler. The basis of deriving this law here is the equation of orbital velocity. We will derive the equation for Kepler’s 3rd Law using the concept of Period of Revolution. In this process, the equation of Time Period Of Revolution of earth satellite would be derived as well. We’ll also solve sample numerical problem here using this law. So let’s start!
Time Period of Revolution of the Satellite about the earth and Kepler’s Third Law derivation
Say, Period of revolution is T and the radius of the orbit is r. And r = R + h = Radius of the earth + height of the satellite from the surface of the earth.
Time Period of revolution of earth satellite
So period of revolution = T = (2 π r) /V, where V is the Orbital Velocity of the satellite. ……………………….. (1)
From equation 1 and 2, we get,
T = (2 π r)/V = [(2 π r)] /[R (g/r)^(1/2)] =[ 2π / R]. √[r ^3/g]
Here we get an important equation for Period of Revolution of earth satellite:
T = [ 2π / R]. √[r ^3/g] ……………………………(3) here R is the radius of the earth and r = R + h
In the next section we will use the above equation to derive Kepler’s Third Law
Derivation of Kepler’s Third Law
Now if we square both side of equation 3 we get the following:
T^2 =[ (4 . π^2)/(R^2)]. r^3/g ……………………………(4)
Here, (4. π^2)/(R^2) and g are constant as the values of π (Pi), g and R are not changing with time.
So we can say, T^2 ∝ r ^3. …………….. (5) .[Kepler’s Third Law equation]
Here this Kepler’s Third Law equation says that square of the Orbital Period of Revolution is directly proportional to the cube of the radius of the orbit.
If the orbit is not circular in the truest sense and rather elliptical, then this law states like this,
square of the Orbital Period of Revolution varies with the cube of the semi-major axis of the orbit. (ref:Wiki)
This is known as Kepler’s third Law.
From this equation of Kepler’s Third Law it comes out clearly that the mass of the object in revolution has no effect on the Period of Revolution.
And this law is applicable for the revolution of any planet and satellite.
Now let’s solve a numerical problem using this formula, in the next paragraph.
Kepler’s Third Law – Sample Numerical Problem using Kepler’s 3rd law:
Two satellites Y and Z are rotating around a planet in a circular orbit. The radii of the orbits for Y and Z are 4R and R respectively. Now if the speed of Y satellite is 3v then what is the speed of satellite Z?
We would use the Kepler’s third law here.
Say, radius of the orbit of Y and Z are Ry and Rz respectively.
As given data, Ry = 4R and Rz = R
From the 3rd Law of Kepler,
Ty^2/Tz^2 =Ry^3/Rz^3 = (4R)^3/R^3 = 64. So, Ty/Tz = 8 ……………. (1)
Continuing with this equation (equation 1) and bringing in the orbital velocity in it we get:
(2 π Ry/Vy) / (2 π Rz/Vz) = 8
=> (Ry Vz) /(Rz Vy) = 8
=> (Ry/Rz).(Vz/Vy) = 8
=> (4R/R) .(Vz/Vy) = 8
=> (Vz/Vy) = 8/4 = 2
=> Vz = 2. Vy = 2. 3v = 6v
So the speed of satellite Z is 6v (answer)
Period of revolution of earth satellite – Numerical Problem
Q1) What is the value of Period of Revolution of earth satellite? ignore the height of satellite above the surface of earth.
Given: gravitational acceleration (g) = 10 m/s^2. Radius of earth = R = 6400 km. Take π as 3.14
Period of revolution = T = [ 2π / R]. √[r ^3/g]
Here r = R + h. As h is to be ignored according to the question, so r =R here.
So T = 2 π √(R/g) ……………. (1)
Given: R= 6400 km = 6400 X 10^3 meter. g = 10 m/s^2
Putting the values given, we get,
T = 2 X 3.14 X √ [(6400X10^3)/10] = 2 X 3.14 X 800 = 5024 seconds = 83.73 minutes.